Source Transformation

Transform the practical voltage source. The 24 V in series with 6 Ω becomes a current source of

Its arrow points into node \(A\), because that is the terminal the source's \(+\) faced.

Now everything is in parallel. The two current sources aid each other and simply add; the two resistors merge:

Read off the answer:

Check by nodal analysis on the original circuit:

Notice what the conversion bought. Before it, the circuit had a voltage source and a current source that could not be combined at all. After it, three parallel branches merged in one line. The transformation did no arithmetic of its own — it rearranged the network so that the arithmetic became possible.

The voltage form. A source \(V_s\) in series with \(R\), delivering current \(I\) at terminal voltage \(V\), obeys KVL:

The current form. A source \(I_s\) in parallel with \(R\) obeys KCL at its terminal — the source current splits between the shunt and the load:

The two agree for every \(I\) precisely when their constant terms match:

Both relations are then the same straight line in the \((V, I)\) plane — Set 10, Problem 2's line, with intercepts \(V_s\) and \(I_s\) and slope \(-1/R\). Two points fix a line, and the two forms share both.

What "indistinguishable" covers. Only the terminal pair \((V, I)\). Anything connected outside — linear or not — behaves identically, because it sees only that relation.

What it does not cover. Everything internal. The current in \(R\) differs between the two forms, and so does the power in it and in the source. Problem 10 measures the difference on Problem 1's circuit: 48 W supplied in one form, 144 W in the other.

Which theorem this is. Thévenin's and Norton's, applied to a single branch rather than a whole network:

Source transformation is therefore not an independent result. It is the smallest possible case of Sets 9 and 10 — and, run in the other direction, it is how those theorems are often applied in practice.

Step 1 — convert the current source with its parallel 3 Ω into a voltage source:

The 3 Ω has moved from being a shunt at \(A\) to being in series with the source — which is exactly what makes the next step possible.

Step 2 — merge. That 3 Ω is now in series with the connecting 3 Ω:

The circuit is now a single loop: 18 V behind 6 Ω, driving the 6 Ω at \(B\):

Check by nodal analysis on the original:

The second gives \(V_A = 1.5V_B\); substituting into the first yields \(V_B = 9\ \text{V}\) and \(V_A = 13.5\ \text{V}\;\checkmark\)

A warning about \(V_A\). The transformed circuit has no node \(A\) at all — it was absorbed into the merged 6 Ω. Any quantity inside the transformed region is lost, so if the question had asked for \(V_A\) or the current in the original 3 Ω shunt, the reduction would have to stop one step earlier.

Step 1. The 2 A source in parallel with the 8 Ω becomes a voltage source:

Because the source draws current out of \(X\), the equivalent 16 V opposes the direction that would push current into \(X\) — which is what makes it aid the 10 V source later.

Step 2. That 8 Ω is now in series with the 4 Ω, so they merge, and the branch converts back:

Step 3. That 12 Ω now sits beside the 6 Ω at node \(Y\), so they merge and the branch converts once more:

The final loop contains the 10 V source, the 3 Ω, the 4 Ω and the \(16/3\) V source, the two sources aiding:

Check by nodal analysis on the original circuit. With the 2 A leaving node \(X\):

Solving gives \(V_Y = 24/7 = 3.43\ \text{V}\), so \(i_a = (10 - V_Y)/3 = 46/21\ \text{A}\;\checkmark\)

Step 1 — convert. The 20 V source with its series 5 kΩ becomes

with its arrow towards the top node, the same sense as the existing 2 mA source.

Step 2 — combine. The two current sources are now in parallel and aid each other:

Step 3 — divide. This 6 mA now feeds \(R_1\) in parallel with \(R_2\), so the current-divider rule applies — with the other resistance in the numerator:

Check. The node voltage is \(6\ \text{mA} \times (5\text{k} \parallel 7.5\text{k}) = 6 \times 3 = 18\ \text{V}\), so

Why the conversion was the right move. Before it, a voltage source and a current source shared a node and could not be combined. Afterwards they were both currents in parallel — the one arrangement in which sources add trivially. Converting towards the form that matches the connection is the general rule: parallel connections want current sources, series connections want voltage sources.

Step 1 — the far branch. The 30 mA source in parallel with 100 Ω becomes a voltage source:

It is now in series with the 8 V source, so the two add to give an 11 V branch behind 100 Ω.

Step 2 — convert both source branches to current form. Each has a 100 Ω in series, so each transforms cleanly:

Step 3 — everything is now in parallel at \(v_a\). Two current sources and three 100 Ω resistors:

Hence

Check by Millman's theorem, treating the three branches as sources of 10 V, 11 V and 0 V behind equal 100 Ω resistances:

With equal resistances Millman reduces to the plain average of the three branch voltages — and \((10+11+0)/3 = 7\).

Convert each branch. Both arrows point towards \(A\), because that is where each source's \(+\) terminal faces:

Combine. Being in parallel and aiding, the sources add and the resistances merge:

The Thévenin form, for comparison:

Millman's theorem confirms it directly: \((18 \times 5 + 10 \times 6)/(6+5) = 150/11\;\checkmark\)

Sanity-check the magnitude. The result lies between 10 V and 18 V, as a conductance-weighted average must. The 5 Ω branch is the stiffer of the two, so it pulls the result below the arithmetic mean of 14 V.

Note what happens with unequal sources. A circulating current flows even with nothing attached at \(A\!-\!B\):

The 18 V source drives it and the 10 V source absorbs it — Set 10, Problem 11.

The magnitudes are unchanged — only the direction of the second arrow reverses:

Now they oppose, so the net source is their difference:

The resistance is unaffected. Deactivating both sources gives the same two resistors in parallel whichever way they were pointing:

Comparing the two cases:

Millman confirms the new terminal voltage: \((18 \times 5 - 10 \times 6)/11 = 30/11 = 2.73\ \text{V}\;\checkmark\)

The circulating current has grown sharply. With the sources opposing, the driving voltage round the internal loop is \(18 + 10 = 28\ \text{V}\) rather than \(18 - 10 = 8\ \text{V}\):

Step 1 — convert the voltage source so it can join the current source:

Step 2 — combine at node \(A\). Both sources are now currents into \(A\), and both resistors are shunts there:

Step 3 — convert back, so the 2 Ω can merge with the 4 Ω in series:

Step 4 — a single loop. 12 V behind 6 Ω, driving the 12 Ω:

Check by nodal analysis on the original two-node circuit:

The second gives \(V_A = \tfrac43 V_B\); substituting reduces the first to \(\tfrac34 V_B = 6\), so \(V_B = 8\ \text{V}\) and \(V_A = 32/3 = 10.67\ \text{V}\;\checkmark\)

Count the work. Four transformations, each a multiplication or a parallel combination, against two simultaneous equations. Neither is long here — but the transformation route scales to any number of stages without the system of equations growing.

Both forms give \(V_A = 24\ \text{V}\). Now look inside each.

The original circuit. The 24 V source drives its 6 Ω into a node that also sits at 24 V, so the current in that branch is

The voltage source delivers nothing, and its series resistor dissipates nothing. The entire circuit is fed by the 2 A source.

The transformed circuit. Here the 6 Ω is a shunt with the full 24 V across it, so it carries 4 A and the 4 A source is fully loaded:

The external element agrees exactly; nothing else does. The transformed circuit appears to burn three times as much power, and its 6 Ω resistor would need a 100 W rating where the real one dissipates nothing at all.

Why this is not a contradiction. Problem 2 proved equivalence at the terminals only. The transformation is a statement about the pair \((V, I)\) at the branch's terminals, and it never claimed anything about what happens inside:

The practical rule. Transform freely to find a current or voltage. Then, for any question about power, efficiency, component rating or fault current, return to the original circuit and use the answer you found.

Case 1 — an ideal voltage source, with no series resistance. The conversion would need

This is Set 10, Problem 14 restated: no Norton form exists. Instead: use a supernode (Set 7), or note that the source simply fixes a node voltage.

Case 2 — an ideal current source, with no parallel resistance:

Instead: use a supermesh (Set 5), or note that the source fixes a mesh current.

Case 3 — the resistance is there but in the wrong place. A resistor in parallel with a voltage source, or in series with a current source, cannot be used for the conversion. Problem 12 works this case in detail.

Case 4 — no element is in simple series or parallel with anything. A bridge has no starting move at all, however many sources it contains. Problem 13 examines this.

Summarising:

A useful observation about the first two cases. A real source always has some internal resistance, so in principle the transformation is always available. It is only the idealisation that blocks it — and the same idealisation is what forces supernodes and supermeshes to exist at all.

Establish the truth first, without any transformation. The terminals see a divider:

The correct transformation uses the series resistor, because that is the one the source's own current must pass through:

The 6 Ω is untouched by the conversion — it was already a shunt and remains one.

Now the two resistors are both shunts and merge:

The wrong transformation uses the parallel 6 Ω:

Half as large again as the truth — and note that it reproduces the source's own 12 V, which is the tell-tale sign: the 3 Ω has vanished from the calculation entirely.

The test to apply. Ask which resistor carries the source's whole current:

Equivalently: the resistor that would remain if you deactivated the source and looked into its own two terminals.

The first move. The 8 V source is ideal — no series resistance is given — so it cannot be transformed at all. That alone stops the method (Problem 11, Case 1).

Suppose we grant it a series resistance and convert it anyway. The method now needs the resulting shunt resistance to merge with something. It cannot:

The reason, in the language of Set 8. Two elements are in series only if they share a node of degree two, and in parallel only if they share both nodes. The bridge graph is \(K_4\) — every node has degree three, and no two nodes are joined by more than one branch. Neither condition can be met anywhere:

Set 8, Problem 1 noted this as the reason the bridge could not be reduced by series–parallel combination. It is the same obstruction here.

What works instead. Every method that does not require a series or parallel pair:

Thévenin's is the interesting one: removing the 18 Ω arm breaks the bridge into two independent dividers, and the obstruction disappears. Deletion succeeds where transformation cannot start.

The transformation is legal here, because the dependent source has a genuine series resistance and its controlling variable \(v_x\) lives outside the branch being transformed:

The dependent source keeps its dependence; only its form changes.

Now KCL at \(A\), where \(v_x = V_A\). Two current sources inject, two resistors shunt:

Collecting terms:

Check without transforming. KCL at \(A\) with the dependent source left in voltage form:

When it is not legitimate. If the controlling variable is the voltage across, or the current through, the very resistor being moved, then the transformation destroys the quantity the source depends on. Before converting, check that the controlling variable survives the move — express it in terms of quantities outside the transformed branch if necessary.

Take \(n\) branches, the \(k\)th being a source \(V_k\) in series with \(R_k\), all connected across the same terminal pair. Transform every branch:

All the current sources are now in parallel and add; all the conductances add too:

Convert back to get the terminal voltage:

That is Millman's theorem, derived in three lines with no circuit analysis whatever.

Verifying on Problem 7 (18 V behind 6 Ω, 10 V behind 5 Ω):

And on Problem 6, whose three branches are 10 V, 11 V and 0 V behind equal 100 Ω resistances:

With equal resistances the conductances cancel and Millman collapses to the plain average of the branch voltages — which is why Problem 6 came out at exactly 7 V.

Reading the formula. It is a conductance-weighted average, so the result always lies between the largest and smallest \(V_k\), and the stiffest branch (smallest \(R_k\)) dominates. A branch with \(V_k = 0\) — a plain resistor — still counts in the denominator, dragging the result towards zero.

Thévenin at \(B\), with the 12 Ω removed. The open-circuit voltage is the voltage at \(A\), since no current flows in the 4 Ω. KCL at \(A\):

The Thévenin resistance. Short the 12 V source and open the 2 A source:

Reconnect the load:

Now compare the two routes step by step:

Identical intermediate quantities throughout — 2 Ω, 12 V, 6 Ω — reached by different names.

Why they coincide. Problem 2 showed source transformation is Thévenin–Norton applied to a single branch. Applying it repeatedly therefore constructs the Thévenin equivalent one element at a time, and the last conversion before the load simply names the result.

Set 8, Problem 12 fixed the counts by topology alone:

Every merge that a transformation enables removes a node or a branch, so both counts fall.

What each operation does:

The conversion alone changes nothing — it only creates the opportunity. The reduction comes from the merge that follows.

Worked on Problem 19's circuit. As given it has nodes \(A\), \(B\) and the reference, so nodal analysis needs two equations. After the transformations of that problem it is a single loop:

When this is the best use of the method. On a large circuit that is partly reducible: transform the ladder-like regions to shrink them, then apply nodal or mesh analysis to whatever irreducible core remains. Neither method alone would be as quick.

The caution from Problem 3. Each merge destroys the interior of the region it absorbs. If the question asks for a current inside a merged group, that merge must not be made — so reduce everything except the region of interest.

The rule. The current source's arrow points towards the terminal at which the voltage source's \(+\) sign was. Reversing the voltage source reverses the arrow.

The justification. Both forms must give the same short-circuit current. Shorting the voltage form drives

so the current form must deliver the same current in the same direction — which fixes the arrow. There is no convention involved; it is forced.

The worked comparison. Take a 20 V source behind 4 Ω feeding node \(N\), with a 3 A source also injecting into \(N\) and a 4 Ω resistor to the reference.

With the arrow reversed — the whole of the error:

Not merely a wrong magnitude but a sign reversal, and no step of the subsequent arithmetic would look suspicious.

Check the correct answer directly:

A cheap safeguard. After transforming, ask what the terminal voltage would be with nothing connected. It must equal the original \(V_{oc}\). Here \(5\ \text{A} \times 4\ \Omega = 20\ \text{V}\), matching the source — while the reversed version gives \(-20\ \text{V}\) and announces the error immediately.

Step 1 — the left source, so it can join the 2 A source:

Step 2 — combine at \(A\):

Step 3 — convert back and absorb the connecting 3 Ω:

Everything left of \(B\) is now a single practical source: 24 V behind 6 Ω.

Step 4 — convert both branches at \(B\) to current form so they can be added:

Step 5 — combine and divide:

Check by nodal analysis, which needs two simultaneous equations:

Solving gives \(V_A = 16.5\ \text{V}\) and \(V_B = 9\ \text{V}\), so \(I_{6\Omega} = 1.5\ \text{A}\;\checkmark\)

What it is. Thévenin's and Norton's theorems applied to a single branch (Problem 2), used repeatedly. It is not an independent result and needs no separate proof.

What it guarantees. Exact equality of the terminal pair \((V, I)\) of the transformed branch — and nothing else. Internal currents, internal powers and efficiency are all destroyed (Problem 10: 48 W becomes an apparent 144 W).

Its place among the methods.

It is the cheapest method and the most easily blocked — the exact trade the table records.

The decision procedure. Before reaching for it, ask three questions in order:

Three yeses and the method will work and will be fastest. A no to (1) or (2) means it cannot start (Problems 11, 13); a no to (3) means it can start but will destroy the answer (Problem 3).

What it produces as a by-product. Millman's theorem falls out in three lines (Problem 15), the Thévenin equivalent is constructed incrementally (Problem 16), and a partly-reducible network can be shrunk before nodal analysis (Problem 17). None of these is the method's stated purpose, and all three are worth more than the arithmetic saving.